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\newcommand{\CourseName}{高等代数一考试A}
\newcommand{\CourseClass}{2023级数学与应用数学、应用统计学}
\newcommand{\CourseDate}{2024年1月}

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\begin{document}

\maketitle

\thispagestyle{fancy} % 第一页也显示“第几页，共几页”的信息。

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%% 题目和解答从这里开始

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\begin{enumerate}

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%\newpage 
\item  %选择题第1题
行列式 $\left|\begin{array}{cccc}8 & 27 & 64 & 125 \\ 4 & 9 & 16 & 25 \\ 2 & 3 & 4 & 5 \\ 1 & 1 & 1 & 1\end{array}\right|$
的值等于多少？ \dotfill (\qquad)

\begin{tasks}(4) %每行4个选项
\task [A.]  $12$
\task [B.]  $-12$
\task [C.]  $16$
\task [D.]  $-16$
\end{tasks}

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%%选择题第1题：解答
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{\color{red} 解答：A. 使用初等行变换，化为上三角行列式。先通过交换两行，将数字小的行换到上面。

}
\fi

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%\newpage 
\item  %选择题第2题
设$\boldsymbol{\eta}_{1}, \boldsymbol{\eta}_{2}$ 为线性方程组 $\boldsymbol{Ax}=\boldsymbol{\beta}$的解, $\boldsymbol{A}$ 的秩为$2$, $\boldsymbol{\eta}_{1}-\boldsymbol{\eta}_{2}=(1,0,2)^{\mathrm{T}}, \boldsymbol{\eta}_{1}+\boldsymbol{\eta}_{2}=(1,-2,2)^{\mathrm{T}}$, 则对任意常数 $k$, 方程组 $\boldsymbol{A x}=\boldsymbol{\beta}$ 的通解是什么？  \dotfill (\qquad)

\begin{tasks}(2) %每行2个选项
\task [A.]  $(1,0,2)^{\mathrm{T}}+k(1,-2,2)^{\mathrm{T}}$.
\task [B.]  $(1,-1,2)^{\mathrm{T}}+k(2,0,4)^{\mathrm{T}}$.
\task [C.]  $(0,1,0)^{\mathrm{T}}+k(1,-2,2)^{\mathrm{T}}$.
\task [D.]  $(1,-2,2)^{\mathrm{T}}+k(1,0,2)^{\mathrm{T}}$. 
\end{tasks}

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%%选择题第2题：解答
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{\color{red} 解答：B. 根据解向量的长度可知未知数个数 $n=3$. 
根据 $\boldsymbol{\eta}_{1}, \boldsymbol{\eta}_{2}$ 为线性方程组 $\boldsymbol{Ax}=\boldsymbol{\beta}$的解可知 
$\boldsymbol{\eta}_{1}-\boldsymbol{\eta}_{2}=(1,0,2)^{\mathrm{T}}$ 为齐次线性方程组 $\boldsymbol{Ax}=0$的解。
因为 $n-r(\boldsymbol{A})=1$, 所以这是一个基础解系。
$\boldsymbol{\eta}_{1}=(1,-1,2)^{\mathrm{T}}$ 是线性方程组 $\boldsymbol{Ax}=\boldsymbol{\beta}$的一个特解。

}
\fi

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\item  %选择题第3题
设 $\boldsymbol{A}$ 是三阶方阵, $\boldsymbol{A}^{*}$ 是其伴随矩阵, $\boldsymbol{A}$ 的所有二阶子式都等于零, 则
下述正确的是哪个？  \dotfill (\qquad)

\begin{tasks}(2) %每行2个选项
\task [A.]  $r(\boldsymbol{A})=1, r\left(\boldsymbol{A}^{*}\right)=0$.
\task [B.]  $r(\boldsymbol{A}) \leq 1, r\left(\boldsymbol{A}^{*}\right)=0$.
\task [C.]  $r(\boldsymbol{A})=2, r\left(\boldsymbol{A}^{*}\right)=1$.
\task [D.]  $r(\boldsymbol{A}) \leq 1, r\left(\boldsymbol{A}^{*}\right)=1$. 
\end{tasks}

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%%选择题第3题：解答
\ifnum\showsolution>0

{\color{red} 解答：B. 根据矩阵的秩的定义，若 $r(\boldsymbol{A})=r$, 则矩阵 $A$ 中存在 $r$ 子式不等于零，而且任意 $r+1$ 阶子式都等于零。所以根据 $\boldsymbol{A}$ 的所有二阶子式都等于零可知 $r(\boldsymbol{A})\le 1$. 
又根据伴随矩阵的定义，$\boldsymbol{A}^*$ 是矩阵 $\boldsymbol{A}$ 的代数余子式排列得到，所以这里的 $\boldsymbol{A}^*$ 是零矩阵。
}
\fi

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%\newpage 
\item  %选择题第4题
已知任一 $n$ 维向量均可由向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n}$ 线性表示, 则关于这个向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n}$ 的下述说法中，正确的是哪个？ \dotfill (\qquad)

\begin{tasks}(4) %每行4个选项
\task [A.]  线性相关
\task [B.]  秩等于 $n$
\task [C.]  秩小于 $n$
\task [D.]  以上都不对 
\end{tasks}

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%%选择题第4题：解答
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{\color{red} 解答：B. 此时向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n}$ 
与向量组 
$$
\boldsymbol{\varepsilon}_{1}=(1,0,\cdots,0)^{\mathrm{T}}, 
\boldsymbol{\varepsilon}_{2}=(0,1,\cdots,0)^{\mathrm{T}}, 
\cdots, 
\boldsymbol{\varepsilon}_{n}=(0,0,\cdots,1)^{\mathrm{T}}
$$ 
可以相互线性表示。因此这个向量组的秩等于 $n$. 


}
\fi

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%\newpage 
\item  %选择题第5题
设 $\boldsymbol{A}, \boldsymbol{B}$ 为正定实对称矩阵, 则 下述说法中，正确的是哪个？ \dotfill (\qquad)

\begin{tasks}(1) %每行1个选项
\task [A.]  $\boldsymbol{A B}, \boldsymbol{A}+\boldsymbol{B}$ 都一定是正定矩阵
\task [B.]  $\boldsymbol{A B}$ 一定是正定矩阵, $\boldsymbol{A}+\boldsymbol{B}$ 不一定是正定矩阵
\task [C.]  $\boldsymbol{A B}, \boldsymbol{A}+\boldsymbol{B}$ 都不一定是正定矩阵
\task [D.]  $\boldsymbol{A B}$ 不一定是正定矩阵, $\boldsymbol{A}+\boldsymbol{B}$ 一定是正定矩阵
\end{tasks}

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%%选择题第5题：解答
\ifnum\showsolution>0

{\color{red} 解答：D. 可以找到两个正定对称矩阵 $A,B$ 的例子，使得 $AB\neq BA$, 此时 $AB$ 不是对称矩阵。
用实二次型考虑，可知若 $X^{\mathrm{T}}AX$ 和 $X^{\mathrm{T}}BX$ 都是正定二次型，则 $X^{\mathrm{T}}(A+B)X$ 也是正定二次型。

}
\fi

%选择题结束
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\item  %填空题第1题
排列$246 \cdots (2n-4)(2n-2)(2n)(2n-1)(2n-3) \cdots 531$的逆序数是$\underline{\qquad \qquad}$.

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%%填空题第1题：解答
\ifnum\showsolution>0

{\color{red} 解答：$n^2$. 
考虑 $n=4$, 这时的排列为 $2,4,6,8,7,5,3,1$
从左到右写出所有的逆序，可得
\begin{equation*}
\begin{aligned}
21, \\ 
43, 41, \\
65, 63, 61, \\ 
87, 85, 83, 81, \\
75, 73, 71, \\
53, 51, \\
31. 
\end{aligned}
\end{equation*}

可见逆序的个数为 $1+2+3+4+3+2+1=16=(4)^2.$
容易推知一般情况共有 $n^2$ 个逆序。
}

\fi

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%\newpage 
\item  %填空题第2题
设$\boldsymbol{A}$ 是 $m \times n$ 矩阵, 若 $r(\boldsymbol{A})=m$, 则非齐次线性方程组 $\boldsymbol{Ax}=\boldsymbol{\beta} \underline{\qquad \qquad}$有解(填写: 一定或不一定).

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\ifnum\showsolution>0

{\color{red} 解答：一定。
%因为 $r(\boldsymbol{A})=m$, 所以矩阵 $\boldsymbol{A}$ 中一定有 $m$ 阶子式且值不等于零，所以 $n\ge m$. 
因为增广矩阵 $\overline{\boldsymbol{A}}=(\boldsymbol{A},\boldsymbol{\beta})$ 是 $m\times (n+1)$ 矩阵，所以其秩最多也是 $m$. 所以 $r(\overline{\boldsymbol{A}})=r(\boldsymbol{A})$. 
根据线性方程组有解的判别定理，$\boldsymbol{Ax}=\boldsymbol{\beta}$ 一定有解。

}

\fi

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\item  %填空题第3题
设$3$阶行列式$|\boldsymbol{A}|$ 的值为 $\dfrac{1}{2}$, $\boldsymbol{A}^*$是 $\boldsymbol{A}$ 的伴随矩阵, 则$|(2\boldsymbol{A})^{-1}-2\boldsymbol{A}^*|=\underline{\qquad \qquad}$.

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%%填空题第3题：解答
\ifnum\showsolution>0

{\color{red} 解答：$-\frac{1}{4}$. 根据 $\boldsymbol{A}^{-1} = \frac{1}{|\boldsymbol{A}|}\boldsymbol{A}^*$ 可得 
$\boldsymbol{A}^* = |\boldsymbol{A}| \boldsymbol{A}^{-1}$. 因此 
\begin{equation*}
\begin{aligned}
\left\lvert (2\boldsymbol{A})^{-1}-2\boldsymbol{A}^* \right\rvert 
&= \left\lvert \frac{1}{2}\boldsymbol{A}^{-1}-2|\boldsymbol{A}|\boldsymbol{A}^{-1} \right\rvert 
= \left\lvert \frac{1}{2}\boldsymbol{A}^{-1}-\boldsymbol{A}^{-1} \right\rvert \\ 
&= \left\lvert -\frac{1}{2}\boldsymbol{A}^{-1} \right\rvert 
= \left(-\frac{1}{2}\right)^3 \left\lvert\boldsymbol{A}^{-1} \right\rvert 
= -\frac{1}{8}\cdot 2 = -\frac{1}{4}. 
\end{aligned}
\end{equation*}

}

\fi

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%\newpage 
\item  %填空题第4题
已知 $n$ 阶方阵 $\boldsymbol{A}, \boldsymbol{B}$ 满足 $\boldsymbol{A}+\boldsymbol{B}=\boldsymbol{E}, \boldsymbol{AB}=\boldsymbol{O}$, 则 $r(\boldsymbol{A})+r(\boldsymbol{B})\underline{\qquad}n$(填写:$>$, $<$ 或$=$).

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%%填空题第4题：解答
\ifnum\showsolution>0

{\color{red} 解答：$=$. 
根据 $\boldsymbol{AB}=\boldsymbol{O}$ 可知矩阵 $\boldsymbol{B}$ 的每个列向量都是齐次线性方程组 $\boldsymbol{Ax}=0$ 的解，所以 $r(\boldsymbol{B})\le n-r(\boldsymbol{A})$, 所以 $r(\boldsymbol{A})+r(\boldsymbol{B})\le n$. 
另一方面，根据 $\boldsymbol{A}+\boldsymbol{B}=\boldsymbol{E}$ 可知 $n = r(\boldsymbol{E})=r(\boldsymbol{A+B})\le r(\boldsymbol{A}) + r(\boldsymbol{B})$. 

}

\fi

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%\newpage 
\item  %填空题第5题
设秩为 $n$ 的 $n$ 元实二次型 $f$ 和 $-f$ 合同, 则 $f$ 的符号差等于 $\underline{\qquad \qquad}$.

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%%填空题第5题：解答
\ifnum\showsolution>0

{\color{red} 解答：$0$.
设这个实二次型的规范形为 $f=z_1^2+\cdots+z_p^2-z_{p+1}^2-\cdots-z_{p+q}^2$, 
则 $-f=-z_1^2 - \cdots-z_p^2 + z_{p+1}^2 + \cdots + z_{p+q}^2$. 
因此 $f$ 的正惯性指数等于 $-f$ 的负惯性指数，而 $f$ 的负惯性指数等于 $-f$ 的正惯性指数。
根据题目条件，$f$ 和 $-f$ 合同，根据惯性定理，$f$ 和 $-f$ 的正、负惯性指数对应相等。
所以只有 $p=q$. 所以$f$ 的符号差等于 $p-q=0$. 

}

\fi

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%填空题结束

\item %计算题第1题
设$\boldsymbol{A}$为$n$阶方阵, 若有$n$阶方阵$\boldsymbol{B}$, 使得$\boldsymbol{AB}=\boldsymbol{BA}=\boldsymbol{E}$, 称$\boldsymbol{A}$为可逆的. 请叙述8条"矩阵$\boldsymbol{A}$可逆"的充分必要条件(要求: 每一条都不相同).

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%%计算题第1题：解答
\ifnum\showsolution>0

{\color{red}解答：
下列均为$\boldsymbol{A}_{n \times n}$可逆的充要条件: 
\begin{enumerate}[label={(\arabic*)}]

\item  $\boldsymbol{A}_{n \times n}^{\rm{T}}$ 可逆;  %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{A}$ 非退化(或非奇异);  %$\textcolor{red}{(+1)}$

\item  $|\boldsymbol{A}| \neq 0$; %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{Ax}=\boldsymbol{0}$ 只有零解;  %$\textcolor{red}{(+1)}$

\item  对任一$n$维向量 $\boldsymbol{\beta}$, $\boldsymbol{Ax}=\boldsymbol{\beta}$ 有唯一解;  %$\textcolor{red}{(+1)}$

\item  对任一$n$维向量 $\boldsymbol{\beta}$, $\boldsymbol{Ax}=\boldsymbol{\beta}$ 都有解;  %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{A}$的秩等于$n$;  %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{A}$的行秩等于$n$ (或$\boldsymbol{A}$ 行满秩); %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{A}$的列秩等于$n$ (或$\boldsymbol{A}$ 列满秩); %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{A}$ 的行向量组线性无关; %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{A}$ 的列向量组线性无关; %$\textcolor{red}{(+1)}$

\item  任一 $n$ 维向量可由 $\boldsymbol{A}$ 的行向量组线性表出; %$\textcolor{red}{(+1)}$

\item  任一 $n$ 维向量可由 $\boldsymbol{A}$ 的列向量组线性表出; %$\textcolor{red}{(+1)}$

\item  $n$ 维单位行向量组$\boldsymbol{\varepsilon}_1, \boldsymbol{\varepsilon}_2, \cdots, \boldsymbol{\varepsilon}_n$可由 $\boldsymbol{A}$ 的行向量组线性表出; %$\textcolor{red}{(+1)}$

\item  $n$ 维单位列向量组$\boldsymbol{e}_1, \boldsymbol{e}_2, \cdots, \boldsymbol{e}_n$可由 $\boldsymbol{A}$ 的列向量组线性表出; %$\textcolor{red}{(+1)}$

\item  $n$ 维单位行向量组$\boldsymbol{\varepsilon}_1, \boldsymbol{\varepsilon}_2, \cdots, \boldsymbol{\varepsilon}_n$与$\boldsymbol{A}$ 的行向量组等价; %$\textcolor{red}{(+1)}$

\item  $n$ 维单位列向量组$\boldsymbol{e}_1, \boldsymbol{e}_2, \cdots, \boldsymbol{e}_n$与$\boldsymbol{A}$ 的列向量组等价; %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{A}$ 的(等价)标准形为单位矩阵(或 $\boldsymbol{A}$ 与单位矩阵等价, 或$\boldsymbol{A}$ 可以经过一系列初等变换化为单位矩阵); %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{A}$ 能表示成一些(或“有限个”)初等矩阵的乘积; %$\textcolor{red}{(+1)}$

\item  $\boldsymbol{A}$ 的伴随矩阵$\boldsymbol{A}^*$可逆(再根据如上条件又可写出一系列充要条件, 比如$\boldsymbol{A}^*$ 的伴随矩阵$\boldsymbol{A}^{**}$可逆, 如此进行下去, 可写出无穷多充要条件).  %$\textcolor{red}{(+1)}$
 
  \ifnum\showsolution=2 
  \dotfill (\textrm{ \, 每条1分\,  }) 
  \fi

\end{enumerate} 

}

\vspace{0.2cm}

\fi

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%\newpage 
\item %计算题第2题
求下述 $n(n>1)$ 阶方阵 $\boldsymbol{A}$ 的逆阵：
$$
\boldsymbol{A}=\left(\begin{array}{cccccc}
1 & 1 & 1 & \cdots & 1 & 1 \\
0 & 1 & 1 & \cdots & 1 & 1 \\
0 & 0 & 1 & \cdots & 1 & 1 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & 1 & 1 \\
0 & 0 & 0 & \cdots & 0 & 1
\end{array}\right) .
$$

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%%计算题第2题：解答
\ifnum\showsolution>0

{\color{red}解答：
初等变换法: 进行初等行变换 $\left(\boldsymbol{A}, \boldsymbol{E}\right)\xrightarrow[i=1, 2, \cdots, n-1]{r_i-r_{i+1}}\left(\boldsymbol{E} , \boldsymbol{A}^{-1}\right)$, 即得
\[\boldsymbol{A}^{-1}=\left(\begin{array}{cccccc}
  1 & -1 & 0 & \cdots & 0 & 0 \\
  0 & 1 & -1 & \cdots & 0 & 0 \\
  0 & 0 & 1 & \cdots & 0 & 0 \\
  \vdots & \vdots & \vdots & & \vdots & \vdots \\
  0 & 0 & 0 & \cdots & 1 & -1 \\
  0 & 0 & 0 & \cdots & 0 & 1
  \end{array}\right). \] 
  
%(写出初等变换方法$\textcolor{red}{(+4)}$, 算出正确结果$\textcolor{red}{(+4)}$) 
  \ifnum\showsolution=2 
  \dotfill (\textrm{ \, 写出初等变换方法给4分, 算出正确结果给4分\,  }) 
  \fi

}

\vspace{0.2cm}

\fi

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%\newpage 
\item %计算题第3题
讨论参数 $\lambda$为何值时, 下列方程组无解, 有唯一解, 有无穷多解? 在有解时, 利用其导出组的基础解系写出全部的解.
\[\left\{\begin{array}{l}
2 x_{1}+3 x_{2}+x_{3}+x_{4}=1, \\
x_{1}+2 x_{2}-x_{3}+4 x_{4}=2, \\
x_{1}+3 x_{2}-4 x_{3}+11 x_{4}=\lambda .
\end{array}\right.\]

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%%计算题第3题：解答
\ifnum\showsolution>0

{\color{red}解答：
对增广矩阵作初等行变换:
  \[\begin{aligned}
 \overline{\boldsymbol{A}}= (\boldsymbol{A},\boldsymbol{\beta})=\left(\begin{array}{ccccc}
  2 & 3 & 1 & 1 & 1 \\
  1 & 2 & -1 & 4 & 2 \\
  1 & 3 & -4 & 11 & \lambda
  \end{array}\right) & \rightarrow
  \left(\begin{array}{ccccc}
  1 & 2 & -1 & 4 & 2 \\
  2 & 3 & 1 & 1 & 1 \\
  1 & 3 & -4 & 11 & \lambda
  \end{array}\right) \rightarrow \\
  \left(\begin{array}{ccccc}
  1 & 2 & -1 & 4 & 2 \\
  0 & -1 & 3 & -7 & -3 \\
  0 & 1 & -3 & 7 & \lambda-2
  \end{array}\right) & \rightarrow\left(\begin{array}{ccccc}
  1 & 2 & -1 & 4 & 2 \\
  0 & -1 & 3 & -7 & -3 \\
  0 & 0 & 0 & 0 & \lambda-5
  \end{array}\right) .
  \end{aligned} \]
  %\textcolor{red}{(+2)} 
  \twopoints 
  
  \begin{enumerate}[label={(\arabic*)}]
  
  \item 当 $\lambda \neq 5$ 时, 因为 $r(\boldsymbol{A})< r(\overline{\boldsymbol{A}})$, 所以方程组无解. 
  %$\textcolor{red}{(+2)}$
  \twopoints

  \item 当 $\lambda = 5$ 时, 因为 $r(\boldsymbol{A})= r(\overline{\boldsymbol{A}})<n$, 所以方程组有无穷多解. 此时， 
  
  \[
  \overline{\boldsymbol{A}} = (\boldsymbol{A},\boldsymbol{\beta})  \rightarrow
  \left(\begin{array}{ccccc}
  1 & 0 & 5 & -10 & -4 \\
  0 & -1 & 3 & -7 & -3 \\
  0 & 0 & 0 & 0 & 0
  \end{array}\right) \rightarrow
  \left(\begin{array}{ccccc}
  1 & 0 & 5 & -10 & -4 \\
  0 & 1 & -3 & 7 & 3 \\
  0 & 0 & 0 & 0 & 0
  \end{array}\right). \]
  方程组的一般解如下，其中 $c_1, c_2$ 为任意常数，
  \[\left\{\begin{array}{l}
  x_{1}=-4-5c_1+10c_2, \\
  x_{2}=3+3c_1-7c_2, \\
  x_{3}=c_1, \\
  x_{4}=c_2. 
  \end{array}\right. \]
%
  %\textcolor{red}{(+2)} 
  \twopoints 
  
  将一般解写为向量形式（即用导出组的基础解系写出通解）, 
  \[\left(\begin{array}{c}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right)
  =\left(\begin{array}{c}-4 \\ 3 \\ 0 \\ 0 \end{array}\right)
  +c_{1}\left(\begin{array}{c}-5 \\ 3 \\ 1 \\ 0 \end{array}\right)
  +c_{2}\left(\begin{array}{c}10 \\ -7 \\ 0  \\ 1\end{array}\right), \] 
其中 $c_1, c_2$ 为任意常数. 
 %\textcolor{red}{(+2)}
 \twopoints 

  \item 本题没有唯一解的情况. 
    
  \end{enumerate}              

}

\vspace{0.2cm}

\fi

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%\newpage 
\item %计算题第4题
求 $\lambda$ 的取值范围，使得下述实二次型是正定的：
$$f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)
=\lambda x_{1}^{2}+\lambda x_{2}^{2}+\lambda x_{3}^{2}+x_{4}^{2}+2 x_{1} x_{2}+2 x_{1} x_{3}-2 x_{2} x_{3}. $$

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%%计算题第4题：解答
\ifnum\showsolution>0

{\color{red}解答：
二次型的矩阵为 $\boldsymbol{A}=\left(\begin{array}{cccc}\lambda & 1 & 1 & 0 \\ 1 & \lambda & -1 & 0 \\ 1 & -1 & \lambda & 0 \\ 0 & 0 & 0 & 1\end{array}\right). $
%\textcolor{red}{(+1)}
\onepoint

$\boldsymbol{A}$ 的顺序主子式为 $\left|\boldsymbol{A}_{1}\right|=\lambda$, $\left|\boldsymbol{A}_{2}\right|=\lambda^{2}-1$, $\left|\boldsymbol{A}_{3}\right|=(\lambda +1)^{2}(\lambda - 2)$, $\left|\boldsymbol{A}_{4}\right|=(\lambda +1)^{2}(\lambda - 2)$. 
%$\textcolor{red}{(+4)}$
\fourpoints

要使$f$ 为正定的, 必须有 $\lambda>0, \lambda^{2}-1>0,(\lambda+1)^{2}(\lambda-2)>0$, 
%$\textcolor{red}{(+2)}$ 
\twopoints
%
解得 $\lambda>2$. 
%$\textcolor{red}{(+1)}$
\onepoint

}

\vspace{0.2cm}

\fi

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%\newpage 
\item %计算题第5题
将下述实二次型化为规范形，并写出所作的非退化线性替换：
$$f\left(x_{1}, x_{2}, x_{3}\right)=x_{1}^{2}-3 x_{2}^{2}-2 x_{1} x_{2}+2 x_{1} x_{3}-6 x_{2} x_{3}.$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%计算题第5题：解答
\ifnum\showsolution>0

{\color{red}解答：
可用配方法或初等变换法. 

(1) 配方法: 
\[\begin{aligned}
f\left(x_{1}, x_{2}, x_{3}\right) & =x_{1}^{2}-3 x_{2}^{2}-2 x_{1} x_{2}+2 x_{1} x_{3}-6 x_{2} x_{3} \\
& =\left(x_{1}-x_{2}+x_{3}\right)^{2}-4 x_{2}^{2}-x_{3}^{2}-4 x_{2} x_{3} \\
& =\left(x_{1}-x_{2}+x_{3}\right)^{2}-\left(2 x_{2}+x_{3}\right)^{2}.
\end{aligned}  \]
%\textcolor{red}{(+2)}
\twopoints

令
\[\left\{\begin{array} { l } 
{ y _ { 1 } = x _ { 1 } - x _ { 2 } + x _ { 3 } , } \\
{ y _ { 2 } = 2 x _ { 2 } + x _ { 3 } , } \\
{ y _ { 3 } = x _ { 3 } , }
\end{array}\right.\] 
即 $f\left(x_{1}, x_{2}, x_{3}\right)$ 经非退化线性替换 
\[\left\{\begin{array}{l}
x_{1}=y_{1}+\frac{1}{2} y_{2}-\frac{3}{2} y_{3}, \\
x_{2}=\frac{1}{2} y_{2}-\frac{1}{2} y_{3}, \\
x_{3}=y_{3},
\end{array}\right. \]
%\textcolor{red}{(+2)}
\twopoints
%
可以化为规范形 $y_{1}^{2}-y_{2}^{2}$. 
%$\textcolor{red}{(+2)}$
\twopoints

二次型对应的对称矩阵之间的合同关系为 $\boldsymbol{B}=\boldsymbol{C}^{\mathrm{T}} \boldsymbol{A} \boldsymbol{C}$, 变量的线性替换为 $\boldsymbol{x}=\boldsymbol{C}\boldsymbol{y}$, 其中 
\[\boldsymbol{A}=%\left(\begin{array}{rrr}
\begin{pmatrix}
1 & -1 & 1 \\
-1 & -3 & -3 \\
1 & -3 & 0
\end{pmatrix}, 
%\end{array}\right), %\quad 
\boldsymbol{B}=
\begin{pmatrix}
%\left(\begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 0
\end{pmatrix}, 
%\end{array}\right), %\quad 
\boldsymbol{C}=
\begin{pmatrix}
%\left(\begin{array}{rrr}
1 & \frac{1}{2} & -\frac{3}{2} \\
0 & \frac{1}{2} & -\frac{1}{2} \\
0 & 0 & 1
\end{pmatrix}, 
%\end{array}\right), 
%\quad 
\boldsymbol{x}=
\begin{pmatrix}
%\left(\begin{array}{c}
x_1 \\
x_2 \\
x_3 \\
\end{pmatrix}, 
%\end{array}\right), 
%\quad 
\boldsymbol{y}=
\begin{pmatrix}
%\left(\begin{array}{c}
y_1 \\
y_2 \\
y_3 \\
\end{pmatrix}. \] 
%\end{array}\right).   \]
%\textcolor{red}{(+2)}
\twopoints


(2) 初等变换法: 实二次型$f$的矩阵为上述实对称矩阵 $\boldsymbol{A}$. 构造分块矩阵，并对矩阵 $\boldsymbol{A}$ 进行初等行变换和初等列变换，并将列变换记录在单位矩阵 $\boldsymbol{E}$ 上，
$$\left(\begin{array}{c}\boldsymbol{A} \\ \boldsymbol{E}\end{array}\right)
%\stackrel{\text{合同变换}}{\longrightarrow} 
\xrightarrow[]{\textrm{合同变换}}
\left(\begin{array}{c}\boldsymbol{B} \\ \boldsymbol{C}\end{array}\right),$$ 
其中 $\boldsymbol{B}, \boldsymbol{C}$ 也同上，规范形和非退化线性替换也同上。
}

\vspace{0.2cm}

\fi

%%计算题结束
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\ifnum\showsolution=1 \newpage \fi 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %证明题第1题
设 $\boldsymbol{A}, \boldsymbol{B}$ 分别是 $n \times m$ 和 $m \times n$ 矩阵. 
\begin{enumerate}[label={(\arabic*)}]
\item %(1) 
证明: 
$\left|\begin{array}{cc}
\boldsymbol{E}_{m} & \boldsymbol{B} \\
\boldsymbol{A} & \boldsymbol{E}_{n}
\end{array}\right|=\left|\boldsymbol{E}_{n}-\boldsymbol{A B}\right|;$

\item %(2) 
若$\lambda$为一非零的数, 证明: $\left|\lambda \boldsymbol{E}_{n}-\boldsymbol{A B}\right|=\lambda^{n-m}\left|\lambda \boldsymbol{E}_{m}-\boldsymbol{B A}\right|;$

\item %(3) 
若$r(\boldsymbol{A})=1$, 证明: $\boldsymbol{A}$一定可以写成一个列向量乘以一个行向量;

\item %(4) 
求下述 $n$ 阶行列式的值：
$$\left|\begin{array}{cccccc}
2 & 2 & 3 & 4 & \cdots & n \\ 
1 & 3 & 3 & 4 & \cdots & n \\ 
1 & 2 & 4 & 4 & \cdots & n \\ 
\vdots & \vdots & \vdots & \vdots & & \vdots \\ 
1 & 2 & 3 & 4 & \cdots & n+1
\end{array}\right|.$$

\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%证明第1题：解答
\ifnum\showsolution>0

{\color{red}证明：
\begin{enumerate}[label={(\arabic*)}]

\item %(1) 
根据分块矩阵的乘法可得
\[\left(\begin{array}{cc}
\boldsymbol{E}_{m} & \boldsymbol{O} \\
-\boldsymbol{A} & \boldsymbol{E}_{n}
\end{array}\right)\left(\begin{array}{cc}
\boldsymbol{E}_{m} & \boldsymbol{B} \\
\boldsymbol{A} & \boldsymbol{E}_{n}
\end{array}\right)=\left(\begin{array}{cc}
\boldsymbol{E}_{m} & \boldsymbol{B} \\
\boldsymbol{O} & \boldsymbol{E}_{n}-\boldsymbol{A B}
\end{array}\right). \]
%\textcolor{red}{(+3)}
\threepoints

两边取行列式, 得
\[\left|\begin{array}{cc}
\boldsymbol{E}_{m} & \boldsymbol{B} \\
\boldsymbol{A} & \boldsymbol{E}_{n}
\end{array}\right|=\left|\boldsymbol{E}_{n}-\boldsymbol{A B}\right|. \]
%\textcolor{red}{(+2)}
\twopoints

\item %(2) 
因\[\left(\begin{array}{cc}
\boldsymbol{E}_{m} & \boldsymbol{B} \\
\boldsymbol{A} & \boldsymbol{E}_{n}
\end{array}\right)\left(\begin{array}{cc}
\boldsymbol{E}_{m} & \boldsymbol{O} \\
-\boldsymbol{A} & \boldsymbol{E}_{n}
\end{array}\right)=\left(\begin{array}{cc}
\boldsymbol{E}_{m}-\boldsymbol{B A} & \boldsymbol{B} \\
\boldsymbol{O} & \boldsymbol{E}_{n}
\end{array}\right),\]

两边取行列式, 得
\[\left|\begin{array}{cc}
\boldsymbol{E}_{m}-\boldsymbol{B A} & \boldsymbol{B} \\
\boldsymbol{A} & \boldsymbol{E}_{n}
\end{array}\right|=\left|\boldsymbol{E}_{m}-\boldsymbol{B A}\right|. \]
%\textcolor{red}{(+2)}
\twopoints

所以$\left|\boldsymbol{E}_{n}-\boldsymbol{A B}\right|=\left|\boldsymbol{E}_{m}-\boldsymbol{B A}\right|$. 此式中, 用$\dfrac{1}{\lambda}\boldsymbol{A}$代替$\boldsymbol{A}$, 得
\[\left|\boldsymbol{E}_{n}-\left(\dfrac{1}{\lambda}\boldsymbol{A}\right)\boldsymbol{B}\right|=\left|\boldsymbol{E}_{m}-\boldsymbol{B}\left(\dfrac{1}{\lambda}\boldsymbol{A}\right)\right| .\]

于是
\[\frac{1}{\lambda^{n}}\left|\lambda \boldsymbol{E}_{n}-\boldsymbol{A} \boldsymbol{B}\right|=\frac{1}{\lambda^{m}}\left|\lambda \boldsymbol{E}_{m}-\boldsymbol{B} \boldsymbol{A}\right|,\]
\[\left|\lambda \boldsymbol{E}_{n}-\boldsymbol{A} \boldsymbol{B}\right|=\lambda^{n-m}\left|\lambda \boldsymbol{E}_{m}-\boldsymbol{B} \boldsymbol{A}\right|. \]
%\textcolor{red}{(+3)}
\threepoints

\item %(3) 
因$r(\boldsymbol{A})=1$, 根据行秩的定义，必存在某行(不妨设为第 $i$ 行), 使其他各行都是它的倍数. 
%\textcolor{red}{(+2)}
\twopoints

设$\boldsymbol{A}$的第 $i$ 行为 $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$, 而第 $1$ 行, 第 $2$ 行, $\cdots$, 第 $n$ 行分别是它的 $a_{1}, a_{2}, \cdots, a_{n}$ 倍(这时 $a_{i}=1$ ). 于是
\[\boldsymbol{A}=\left(\begin{array}{cccc}
a_{1} b_{1} & a_{1} b_{2} & \cdots & a_{1} b_{n} \\
a_{2} b_{1} & a_{2} b_{2} & \cdots & a_{2} b_{n} \\
\vdots & \vdots & & \vdots \\
a_{n} b_{1} & a_{n} b_{2} & \cdots & a_{n} b_{n}
\end{array}\right)=\left(\begin{array}{c}
a_{1} \\
a_{2} \\
\vdots \\
a_{n}
\end{array}\right)\left(b_{1}, b_{2}, \cdots, b_{n}\right). \]
%\textcolor{red}{(+3)}
\threepoints

\item %(4) 
方法1: 应用上面的等式 $\left|\boldsymbol{E}_{n}-\boldsymbol{A B}\right|=\left|\boldsymbol{E}_{m}-\boldsymbol{B A}\right|$, 可得 

$\left|\begin{array}{cccccc}
2 & 2 & 3 & 4 & \cdots & n \\ 
1 & 3 & 3 & 4 & \cdots & n \\ 
1 & 2 & 4 & 4 & \cdots & n \\ 
\vdots & \vdots & \vdots & \vdots & & \vdots \\ 
1 & 2 & 3 & 4 & \cdots & n+1
\end{array}\right| 
=\left|\left(\begin{array}{ccccc}
1 & 0 & 0 & \cdots & 0 \\ 
0 & 1 & 0 & \cdots & 0 \\ 
0 & 0 & 1 & \cdots & 0 \\ 
\vdots & \vdots & \vdots & & \vdots \\ 
0 & 0 & 0 & \cdots & 1
\end{array}\right)
+\left(\begin{array}{cccccc}
1 & 2 & 3 & 4 & \cdots & n \\ 
1 & 2 & 3 & 4 & \cdots & n \\ 
1 & 2 & 3 & 4 & \cdots & n \\ 
\vdots & \vdots & \vdots & \vdots & & \vdots \\ 
1 & 2 & 3 & 4 & \cdots & n
\end{array}\right)\right| \\ 
=\left|\boldsymbol{E}_{n}+
\left(\begin{array}{c}1 \\ 1 \\ \vdots \\ 1\end{array}\right)
\left(1, 2, \cdots, n\right)\right|
=\left| \boldsymbol{E}_{n}-\left(\begin{array}{c}-1 \\ -1 \\ \vdots \\ -1\end{array}\right)\left(1, 2, \cdots, n\right)\right| $ 
%\textcolor{red}{(+2)}
\twopoints 
\\
$=\left| \boldsymbol{E}_{1}-\left(1, 2, \cdots, n\right)\left(\begin{array}{c}-1 \\ -1 \\ \vdots \\ -1\end{array}\right)\right|
=1+\sum_{i=1}^{n} i=\dfrac{n^{2}+n+2}{2}. $
%\textcolor{red}{(+3)}
\threepoints

方法2: 因为此行列式的行和均相等, 也可以利用求和法和行列式的性质求解, 

$\left|\begin{array}{cccccc}
2 & 2 & 3 & 4 & \cdots & n \\ 
1 & 3 & 3 & 4 & \cdots & n \\ 
1 & 2 & 4 & 4 & \cdots & n \\ 
\vdots & \vdots & \vdots & \vdots & & \vdots \\ 
1 & 2 & 3 & 4 & \cdots & n+1
\end{array}\right|
=(1+1+2+\cdots+n)\left|\begin{array}{cccccc}
1 & 2 & 3 & 4 & \cdots & n \\ 
1 & 3 & 3 & 4 & \cdots & n \\ 
1 & 2 & 4 & 4 & \cdots & n \\ 
\vdots & \vdots & \vdots & \vdots & & \vdots \\ 
1 & 2 & 3 & 4 & \cdots & n+1
\end{array}\right|  
%\textcolor{red}{(+2)}
\twopoints 
\\
=\dfrac{n^{2}+n+2}{2}\left|\begin{array}{cccccc}
1 & 2 & 3 & 4 & \cdots & n \\ 
0 & 1 & 0 & 0 & \cdots & 0 \\ 
0 & 0 & 1 & 0 & \cdots & 0 \\ 
\vdots & \vdots & \vdots & \vdots & & \vdots \\ 
0 & 0 & 0 & 0 & \cdots & 1
\end{array}\right|=\dfrac{n^{2}+n+2}{2}. $
%\textcolor{red}{(+3)}
\threepoints 
\end{enumerate}

}

\vspace{0.2cm}

\fi

%%证明题结束
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\end{enumerate}

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%%试卷结束
\end{document}





